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A fair coin will be tossed twice and a fair die with sides n

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by hazelnut01 » Tue May 09, 2017 8:21 am
A fair coin will be tossed twice and a fair die with sides numbered 1, 2, 3, 4, 5, and 6 will been rolled twice. What is the probability that at least one head will be tossed and at least one number greater than 1 will be rolled?

A. 35/144

B. 1/2

C. 25/48

D. 9/16

E. 35/48

Source : Kaplan
OA=E
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by [email protected] » Tue May 09, 2017 3:07 pm
Hi ziyuenlau,

Since this question asks for two results to occur (at least one head AND at least one number greater than 1), we need to calculate the probabilities of each individual event and then multiply them together.

The two coin tosses produce (2)(2) = 4 possible outcomes

HH
HT
TH
TT

Of those 4 options, 3 of the 4 are what we "want"

When rolling the two dice, there are (6)(6) = 36 options. Only one of those options is NOT what we want (rolling a 1 on both dice), so 35/36 options are what we "want"

(3/4)(35/36) = 35/48

Final Answer: E

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by Matt@VeritasPrep » Thu May 11, 2017 8:44 pm
P(at least one) = 1 - P(none)

So P(≥1H) = 1 - P(0H) = 1 - 1/2*1/2 = 3/4

and

P(≥1 1+) = 1 - P(0 1+) = 1 - 1/6 * 1/6 = 35/36

Then multiply these together: 3/4 * 35/36, for 35/48.
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by Scott@TargetTestPrep » Fri May 19, 2017 5:12 am
ziyuenlau wrote:A fair coin will be tossed twice and a fair die with sides numbered 1, 2, 3, 4, 5, and 6 will been rolled twice. What is the probability that at least one head will be tossed and at least one number greater than 1 will be rolled?

A. 35/144

B. 1/2

C. 25/48

D. 9/16

E. 35/48
We can use the following equations:

1 = P(at least 1 heads) + P(no heads)

P(at least 1 heads) = 1 - P(no heads)

AND

1 = P(at least 1 number greater than 1) + P(no numbers greater than 1)

P(at least 1 number greater than 1) = 1 - P(no numbers greater than 1)

Let's start with the coin.

P(no heads) = 1/2 x 1/2 = 1/4

P(at least 1 heads) = 1 - 1/4 = 3/4

Next we can determine the dice probability.

P(no numbers greater than 1) = 1/6 x 1/6 = 1/36

P(at least 1 number greater than 1) = 1 - 1/36 = 35/36

Thus, the probability of at least one heads and at least one number greater than 1 is 3/4 x 35/36 = 1/4 x 35/12 = 35/48.

Answer: E

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